If $\bar M A M^T=MAM^\dagger$ for all real $A$, must $M$ be real?

Question

Suppose M is an $n \times n$ complex matrix, if $$\bar M A M^T=MAM^\dagger$$ for all real $n \times n$ matrices $A$ does this imply that $M$ must also be real? Here $\bar M$ denotes the element-wise complex conjugate of $M$, $M^T$ denotes the transpose and $M^\dagger$ denotes the conjugate-transpose. I think that this is true and am unable to find a counter-example.

Answer 1

No. First, notice that your condition is equivalent to saying that $M A M^\dagger$ is real for all real $A.$ Second, suppose that $M$ is diagonal with diagonal elements $m_1, \dotsc, m_n.$ Your condition is then equivalent to saying that $m_i \overline{m_j}$ is real for every $i, j.$ But that is true if $m_i = c_i m_1, $ for some real $c_2, \dotsc, c_n,$ and $m_1$ arbitrary complex.

Answer 2

The condition means that $MAM^\dagger$ is a real matrix. If you write $M=X+iY$ where $X,Y$ are real, your condition amounts to $XAY^T=YAX^T$ for every real $A$, i.e. $Y\otimes X=X\otimes Y$. Thus $X,Y$ are linearly dependent, meaning that $M$ must be a complex multiple of a real matrix.